# Differential equation question (HELP)

1. Sep 17, 2006

### Mathman23

Hi

Given the non-homogenous higher order differential equation

$$u'' - 2 u' - 8 u = 6 - 8t$$

with the intial conditions u(1) = 0 and u'(1) = 1.

Find the solution.

I'm told by my professor that I'm supposed to use the fact

$${u''} + {a_1} u' + {a_0} u = f$$
The characteristic polynomial for this differential equation is supposedly

$$z^2 + {a_1} z + {a_0} z = 0$$

Using this fact I find root of the polynomial of the differential equation above to be $$\lambda _{1,2} = 4,-2$$

The I'm suppose to use the fact that (Existence and Uniqueness theorem of Non-liniear differential equations) if $$\lambda {_1} \neq \lambda {_2}$$

$$u_0 (t) : = \int \limit_{t_0} ^t \frac{e^{\lambda_1 (t-s)} - e^{\lambda_2 (t-s)}}{\lambda_1 - \lambda_2} f(s) ds$$

$$t \in I$$

My solution.

Since lambda 1 and lambda 2 er different then there exist a unique partiqular solution in the form:

$$u(t) = C_1 e^{6-8t} + C_2 e^{-8} = 0$$

By inserting part 1 of the condition into the above, I get that C_1 = C_2 = 0.

initial condition 1 = true.

I differentiate u(t) and get $$u'(t) = -8 e^{-2} C_1 = 1$$

here I get that $$C_1 = \frac{1}{-8{e^{-2}}} = \frac{-e^2}{8}$$ and this proves that u'(1) = 1

Am I on the right track here now?

Sincerley Yours
Fred

Last edited: Sep 17, 2006
2. Sep 17, 2006

### arildno

That would be the solution of the associated HOMOGENOUS problem (i.e, when f=0).
In addition to that, you need a particular solution of your diff. eq, try with the trial function at+b, where a, b are constants to be determined

3. Sep 17, 2006

### Mathman23

Please look at my corrected solution above. Am I on the right track?

Sincerely
Fred

4. Sep 17, 2006

### arildno

Your u(t) is NOT a solution of your original diff eq; it is part of it!

5. Sep 17, 2006

### Mathman23

Okay the I not sure but then full solution is

u(t) = C_1 e^{6-8t} + C_2 e^{-8} = 6 - 8t ??

Sincerley

Fred

6. Sep 17, 2006

### arildno

No, insert in your diff. eq. the trial function:
$$u(t)=C_{1}e^{\lambda_{1}t}+C_{2}e^{\lambda_{2}t}+at+b$$
What equations do you see that "a" and "b" must fulfill in order for u(t) to be a solution?

7. Sep 17, 2006

### Mathman23

a and b must have such values that the solution is equal to the original differential equation ?

/Fred

8. Sep 17, 2006

### arildno

Almost:
a and b must have such values that the trial function is a SOLUTION of the original differential equation.

9. Sep 17, 2006

### Mathman23

by inserting a trail equation u(t) = at^2 + b

I get a = 0 and b = 2.

Does that sound right?

Sincerely Fred

10. Sep 17, 2006

### arildno

No, you should insert at+b:
the second derivative of this is zero, so you end up with the equation for a, b:
$$-2a-8(at+b)=6-8t$$
Thus, we get a=1 to match the first power in t, and to zeroth order in t, we're left with:
-2*1-8b=6, that is b=-1

11. Sep 17, 2006

### Mathman23

Okay then the solution is

$$u(t)=C_{1}e^{\lambda_{1}t}+C_{2}e^{\lambda_{2}t}+ t - 1$$

But what about the lambda values could you give me a hint on how til determain them?

Sincerely

Fred

12. Sep 17, 2006

### arildno

Well, you have already found them as the solutions of the characteristic equation!
Surely you have seen this type of exercises before?

13. Sep 17, 2006

### Mathman23

Yes of course, I have had the flue this last week, so I have been under the weather ;)

Just to recap lambda values are found as the solution of

u(t) = z(t)^2 - 2 z(t) - 8 = 0 ?

Sincerely Fred.

p.s. Talking about existence and uniquness. The exist only one solution for the differential equation. Since the solution polynomial for diff. equation is unique ?

14. Sep 17, 2006

### arildno

Yes.
You now find the the UNIQUE solution of the whole initial value problem by fitting the C's so that the initial conditions are satisfied.

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